A stronger version of paracompactness

Question

Given a topological space $(X,\tau)$, recall that a cover $\mathcal{U}$ of $X$ is locally finite if for every point $x\in \mathcal{U}$ has a neighborhood $U$ that intersects finitely many elements of $\mathcal{U}$. Instead, we call a cover $\mathcal{U}$ finitely intersecting if every member of $\mathcal{U}$ intersect finitely many elements of $\mathcal{U}$.

Recall that $X$ is paracompact if every open cover $\mathcal{U}$ has a locally finite open refinement. Just for the purpose of this question, let us call $X$ strongly paracompact if every open cover $\mathcal{U}$ has a finitely intersecting open refinement.

My question is: does this notion coincide with paracompactness?

If the answer is no, I would be very courious to know more about this property. For example:

• Has strong paracompactness been studied before, and what is the right name for it?

• Is it true that all metrizable second countable spaces are strongly paracompact?

Answer 1

Lemma: Let $(U_{\alpha})_{\alpha\in A}$ be a finitely intersecting open cover of a space $X$. Then there is some partition $P$ of $A$ where $(\bigcup_{\alpha\in R}U_\alpha)_{R\in P}$ is a partition of $X$ into clopen sets and where each $R\in P$ is finite or countable.

Proof: Let $E$ be the smallest equivalence relation on $A$ where if $U_\alpha\cap U_\beta\neq\emptyset$, then $(\alpha,\beta)\in E$. Let $P$ be the partition associated with $E$. Then since $(U_\alpha)_{\alpha\in A}$ is finitely intersecting, each $R\in P$ is finite or countable. Clearly, $(\bigcup_{\alpha\in R}U_\alpha)_{R\in P}$ covers $X$. Furthermore, if $R,S\in P,R\neq S$, then $$(\bigcup_{\alpha\in R}U_\alpha)\cap(\bigcap_{\beta\in S}U_\beta) =\bigcup_{\alpha\in R,\beta\in S}(U_\alpha\cap U_\beta)=\emptyset.$$ Therefore, $(\bigcup_{\alpha\in R}U_\alpha)_{R\in P}$ is a partition into open (and hence clopen) sets. $\square$

Proposition: Every connected strongly paracompact space is Lindelof.

Proof: Let $X$ be a connected strongly paracompact space. Let $\mathcal{U}$ be an open cover of $X$. Let $\mathcal{V}$ be a finitely intersecting open refinement of $\mathcal{U}$. By the above lemma, $\mathcal{V}$ must be countable or finite, so $\mathcal{U}$ has a countable subcover. $\square$